Arithmetic of Complex Numbers

Learning goals

Add and subtract complex numbers, real parts with real parts
Multiply by expanding, then exchange $i^2$ for $-1$
Find a conjugate, and use $z\overline{z} = a^2 + b^2$
Divide by rationalizing with the denominator's conjugate

One new symbol, no new laws

When the previous lesson called $a + bi$ a number, that word carried a promise: the new numbers obey the same laws of algebra as the old ones. Addition is still commutative and associative, multiplication still distributes over addition, and subtraction and division still mean what they always meant, namely adding an opposite and multiplying by a reciprocal. The complex numbers add exactly one fact to that rulebook:

i^2 = -1.

Everything in this lesson is forced by that single stipulation. The working strategy throughout is to treat $a + bi$ as a two-term expression in the symbol $i$ and run the ordinary algebra you have used for years. Then exchange $i^2$ for $-1$ the moment it appears. The question worth asking in advance is whether the answers stay inside the system. If you add, subtract, multiply, or divide two complex numbers, is the result again of the form $(\text{real}) + (\text{real})\,i$ ? Watch for the answer as each operation is built, because it will be yes all four times. That is what makes the complex numbers a self-contained place to do mathematics rather than a notational accident.

Addition and subtraction combine like terms

Add two complex numbers by regrouping, which the commutative and associative laws permit, and then factoring $i$ out of the two imaginary terms, which the distributive law permits:

(a + bi) + (c + di) = (a + c) + (bi + di) = (a + c) + (b + d)\,i.

Real parts add to real parts, and imaginary parts add to imaginary parts, precisely the way $x$ -terms and constant terms keep to themselves when you simplify $(3 + 5x) + (2 + 4x)$ . Subtraction is the same computation after distributing the minus sign through both terms of the second number:

(a + bi) - (c + di) = (a - c) + (b - d)\,i.

Notice what the right sides look like. Since $a + c$ and $b + d$ are real numbers, each result is again $(\text{real}) + (\text{real})\,i$ : adding and subtracting complex numbers can never produce anything but a complex number. Two operations down, two to go.

Worked example 1 Adding and subtracting

Compute $(3 - 5i) + (7 + 2i)$ , then $(4 + 3i) - (6 - 8i)$ .

For the sum, collect the real parts and the imaginary parts separately:

(3 - 5i) + (7 + 2i) = (3 + 7) + (-5 + 2)\,i = 10 - 3i.

For the difference, first distribute the minus sign through the second number, turning $-(6 - 8i)$ into $-6 + 8i$ , and only then collect:

(4 + 3i) - (6 - 8i) = 4 + 3i - 6 + 8i = (4 - 6) + (3 + 8)\,i = -2 + 11i.

The single most common slip in this entire lesson happens in that second computation: subtracting only the $6$ and forgetting that the minus sign also flips the $-8i$ . Distribute first, then combine.

Check your understanding

Compute $(5 - 3i) - (2 + 4i)$ .

3 - 7i

3 + i

7 - 7i

3 + 7i

Check your understanding

Compute $(-4 + 6i) + (9 - 2i)$ .

5 + 4i

5 + 8i

-13 + 4i

13 - 4i

Multiplication is expansion plus one substitution

Multiply two complex numbers the way you multiply any two binomials: distribute every term of the first factor across every term of the second. Nothing about that step is new. The new step comes after, when an $i^2$ is sitting in the result and must be exchanged for $-1$ :

(a + bi)(c + di) = ac + ad\,i + bc\,i + bd\,i^2 = (ac - bd) + (ad + bc)\,i.

Read the middle expression carefully. The four products are exactly the four you would write for $(a + bx)(c + dx)$ . The only difference is the last one, where $bd\,i^2 = -bd$ abandons the imaginary column and joins the real part with its sign flipped. That is the whole novelty of complex multiplication. Do not memorize the boxed-looking formula $(ac - bd) + (ad + bc)\,i$ ; it is faster and safer to expand and substitute each time, letting the formula rebuild itself. And once more the result is $(\text{real}) + (\text{real})\,i$ , so multiplication also stays inside the system.

An area model for the product of 2 + 3i and 4 - 5i. Each cell holds one of the four products from the expansion. Three cells are routine, but the bottom-right cell contains i squared, and the rule i squared = -1 folds it back into the real column. That returned +15 is why the real part of the answer is 23 rather than 8.

Worked example 2 Two products in full

Compute $(2 + 3i)(4 - 5i)$ , then $(3 - 2i)^2$ .

The first product is the one the area model above already lays out: expand term by term to get $8 - 10i + 12i - 15i^2$ , then trade $-15i^2$ for $+15$ and collect real parts with real parts:

(2 + 3i)(4 - 5i) = (8 + 15) + (-10 + 12)\,i = 23 + 2i.

The second product is a square, not a product of two different numbers, so it needs its own expansion. Use the binomial pattern $(x - y)^2 = x^2 - 2xy + y^2$ with $x = 3$ and $y = 2i$ , or simply distribute:

(3 - 2i)^2 = 9 - 12i + 4i^2 = 9 - 12i - 4 = 5 - 12i.

In both computations the expansion was ordinary algebra; the only complex-number moment was trading each $i^2$ for $-1$ before collecting.

Check your understanding

Compute $(3 + i)(2 - 5i)$ .

6 - 13i

11 - 13i

11 - 17i

1 - 13i

The conjugate makes products real

One special product deserves its own name, because it powers everything in the rest of the lesson. The conjugate of $z = a + bi$ is the number

\overline{z} = a - bi,

the same number with the sign of its imaginary part flipped. So $\overline{4 + 7i} = 4 - 7i$ , the conjugate of $-3 - 5i$ is $-3 + 5i$ , and $\overline{6i} = -6i$ . Flipping the sign twice restores the original, so $\overline{\overline{z}} = z$ : conjugation is its own undo. A real number is its own conjugate, since $r = r + 0i$ and flipping the sign of $0$ changes nothing. The converse also holds: $\overline{z} = z$ means $a - bi = a + bi$ , and matching the imaginary parts forces $-b = b$ , so $b = 0$ . A complex number equals its own conjugate exactly when it is real.

Check your understanding

What is the conjugate of $-5 + 3i$ ?

-5 - 3i

5 + 3i

5 - 3i

-5 + 3i

Pair a number with its conjugate and both basic operations produce real answers. The sum is $(a + bi) + (a - bi) = 2a$ , twice the real part, with the imaginary parts canceling. The product is the important one.

The product of conjugates is real and never negative#

Expand the product of $z = a + bi$ and $\overline{z} = a - bi$ like any other product:

(a + bi)(a - bi) = a^2 - ab\,i + ab\,i - b^2 i^2 = a^2 + b^2.

The two middle terms are opposites and cancel, and the last term flips sign because $i^2 = -1$ . Compare this with the difference-of-squares pattern $(x + y)(x - y) = x^2 - y^2$ that you know from factoring: the pattern fires here too, giving $a^2 - (bi)^2 = a^2 - b^2 i^2$ . That last $i^2$ turns the familiar difference of squares into a sum of squares. So

z\,\overline{z} = a^2 + b^2,

a real number built from two squares of real numbers. Each square is at least $0$ , so $z\,\overline{z} \ge 0$ always. In fact it is never merely $0$ unless $z$ is: if $a^2 + b^2 = 0$ with $a, b$ real, neither square can be positive without pushing the sum above $0$ , so both squares are $0$ and $a = 0$ , $b = 0$ . Conversely $z = 0$ gives $0^2 + 0^2 = 0$ . So $z\,\overline{z}$ is strictly positive for every complex number except $z = 0$ itself.

Keep the shape of that result in mind: multiplying any complex number by its conjugate erases $i$ entirely and leaves the nonnegative real number $a^2 + b^2$ , strictly positive whenever $z \ne 0$ . The next lesson will draw $a + bi$ as a point in a plane and read $a^2 + b^2$ off the picture geometrically.

Check your understanding

What is $(1 - 4i)(1 + 4i)$ ?

17

-15

1 - 16i

-17

Conjugation has one more property, useful for the chapter’s final lesson: it does not care whether you conjugate before or after adding or multiplying.

Worked example 3 Conjugate products at work

Compute $(3 + 4i)(3 - 4i)$ and $(1 + 2i)(1 - 2i)$ without a full expansion.

Each is a number times its own conjugate, so the theorem answers immediately with $a^2 + b^2$ :

(3 + 4i)(3 - 4i) = 3^2 + 4^2 = 25, \qquad (1 + 2i)(1 - 2i) = 1^2 + 2^2 = 5.

If you distrust the shortcut, expand the first one: $9 - 12i + 12i - 16i^2 = 9 + 16 = 25$ . The cross terms cancel and the $i^2$ flips the last sign, exactly as the proof said they must. Notice both answers are positive real numbers; an $i$ can never survive this pairing.

Division is rationalizing in disguise

You have solved this problem before, in a different costume. When a denominator held a radical, you multiplied by a well-chosen form of $1$ to clear it:

\frac{1}{3 + \sqrt{2}} = \frac{1}{3 + \sqrt{2}} \cdot \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}.

The conjugate $3 - \sqrt{2}$ was chosen because the difference-of-squares pattern makes the new denominator $3^2 - (\sqrt{2})^2 = 7$ , a rational number. Division of complex numbers is the same trick with the same name. To divide by $c + di$ , multiply numerator and denominator by its conjugate $c - di$ , and the proof above guarantees the new denominator is the real number $c^2 + d^2$ :

\frac{a + bi}{c + di} = \frac{a + bi}{c + di} \cdot \frac{c - di}{c - di} = \frac{(a + bi)(c - di)}{c^2 + d^2}.

Multiplying by $\frac{c - di}{c - di}$ is multiplying by $1$ , so the value never changes; only its costume does. After this move, the numerator is one ordinary complex multiplication, and dividing a complex number by the positive real $c^2 + d^2$ just divides each part, since $\frac{p + qi}{m} = \frac{p}{m} + \frac{q}{m}\,i$ .

Pause on when this is allowed. The move needs $c^2 + d^2 \ne 0$ , and the conjugate proof showed $c^2 + d^2 = 0$ happens only for $c + di = 0$ . So the only forbidden division is division by zero, the same single exception the real numbers always had. Every nonzero complex number can divide, and in particular every nonzero $z = c + di$ has a reciprocal,

\frac{1}{c + di} = \frac{c - di}{c^2 + d^2},

again of the form $(\text{real}) + (\text{real})\,i$ . That settles the closure question asked at the start: all four operations take complex numbers to complex numbers. The system that began as one symbol and one rule supports the full arithmetic that the rational and real numbers enjoy, with nothing missing and nothing extra needed.

Worked example 4 A quotient, start to finish

Compute $\dfrac{3 + 2i}{1 - 4i}$ .

The denominator is $1 - 4i$ , so multiply numerator and denominator by its conjugate $1 + 4i$ :

\frac{3 + 2i}{1 - 4i} \cdot \frac{1 + 4i}{1 + 4i} = \frac{(3 + 2i)(1 + 4i)}{(1 - 4i)(1 + 4i)}.

The denominator is a conjugate product, so it is $1^2 + 4^2 = 17$ with no work. Expand the numerator and spend $i^2 = -1$ :

(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i.

Now divide each part by $17$ :

\frac{3 + 2i}{1 - 4i} = \frac{-5 + 14i}{17} = -\frac{5}{17} + \frac{14}{17}\,i.

The answer is not pretty, but it is exact, and it has the required shape of a real part plus a real multiple of $i$ .

Check your understanding

Compute $\dfrac{2 + i}{3 - i}$ .

\dfrac{1}{2} - \dfrac{1}{2}i

\dfrac{5}{8} + \dfrac{5}{8}i

\dfrac{1}{2} + \dfrac{1}{2}i

5 + 5i

This arithmetic already hints at the chapter’s payoff: the quadratic formula can now run on a negative discriminant instead of stopping there, and the final lesson uses exactly the operations you just practiced to finish the job.

Common mistakes

Leaving an $i^2$ unspent. The expansion $(2 + 3i)(4 - 5i) = 8 - 10i + 12i - 15i^2$ is not finished: the last term must become $+15$ and join the real part. Treating $i^2$ as if it were $1$ , or dropping it, is the most common multiplication error, and it changes $23 + 2i$ into a wrong answer like $-7 + 2i$ .
Subtracting only half of the second number. In $(4 + 3i) - (6 - 8i)$ , the minus sign hits both terms: $4 + 3i - 6 + 8i$ . Forgetting to flip the $-8i$ gives $-2 - 5i$ instead of the correct $-2 + 11i$ .
Calling $(a + bi)(a - bi)$ a difference of squares and stopping at $a^2 - b^2$ . The pattern does fire, but it produces $a^2 - b^2 i^2$ , and the $i^2$ flips the sign: the product of conjugates is the sum $a^2 + b^2$ . That plus sign is the entire reason the conjugate clears denominators.
Multiplying only the denominator by the conjugate. To compute $\frac{a + bi}{c + di}$ you multiply by $\frac{c - di}{c - di}$ , a form of $1$ . Multiplying the bottom alone (or using the numerator’s conjugate) changes the value of the fraction, not just its form.
Flipping the wrong sign when conjugating. The conjugate changes only the imaginary part: the conjugate of $-3 - 5i$ is $-3 + 5i$ , not $3 + 5i$ . The real part never moves.
Stopping short of the form $a + bi$ . An answer like $\frac{5 + 10i}{5}$ or $8 - 10i + 12i - 15i^2$ is unfinished work. Reduce every power of $i$ , combine like terms, and split any real denominator across both parts before calling a computation done.

Key takeaways

Complex numbers obey the ordinary laws of algebra plus the single rule $i^2 = -1$ .
Add and subtract like terms: $(a + bi) \pm (c + di) = (a \pm c) + (b \pm d)\,i$ .
Multiply by expanding, then exchange every $i^2$ for $-1$ : $(a + bi)(c + di) = (ac - bd) + (ad + bc)\,i$ . Re-derive this by expansion rather than memorizing it.
The conjugate of $z = a + bi$ is $\overline{z} = a - bi$ . Then $z\,\overline{z} = a^2 + b^2$ , real and never negative, and $0$ only for $z = 0$ .
Divide by rationalizing: multiply numerator and denominator by the conjugate of the denominator, so the denominator becomes the positive real $c^2 + d^2$ ; in particular $\frac{1}{c + di} = \frac{c - di}{c^2 + d^2}$ .
All four operations return answers of the form $(\text{real}) + (\text{real})\,i$ , and only division by $0$ is forbidden.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Complex numbers (Khan Academy)
2.4 Complex Numbers (OpenStax)
Operations on complex numbers (Purplemath)
Complex numbers (Math is Fun)

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Conjugation respects sums and products, and why complex roots pair up

Conjugation respects sums and products#

Let $z = a + bi$ and $w = c + di$ be any two complex numbers, with $a, b, c, d$ real.

For sums, compute both orders. Conjugating after adding gives $\overline{z + w} = \overline{(a + c) + (b + d)\,i} = (a + c) - (b + d)\,i$ . Adding the conjugates gives $\overline{z} + \overline{w} = (a - bi) + (c - di) = (a + c) - (b + d)\,i$ . The two agree, so $\overline{z + w} = \overline{z} + \overline{w}$ .

For products, the multiplication formula from earlier gives $zw = (ac - bd) + (ad + bc)\,i$ , so conjugating after multiplying yields

\overline{zw} = (ac - bd) - (ad + bc)\,i.

Multiplying the conjugates first instead,

\begin{aligned} \overline{z}\,\overline{w} &= (a - bi)(c - di) = ac - ad\,i - bc\,i + bd\,i^2 \\ &= (ac - bd) - (ad + bc)\,i, \end{aligned}

the same number. So $\overline{zw} = \overline{z}\,\overline{w}$ as well. In words: conjugate first or compute first, the answer is identical, for addition and for multiplication. The answer is therefore identical for anything built out of those two operations, such as subtraction, squaring, and higher powers.

Here is the payoff that property sets up. In the last chapter, the discriminant sorted every quadratic, and the verdict for $\Delta < 0$ was blunt: no real solutions. Take $z^2 - 4z + 5 = 0$ , whose discriminant is $\Delta = (-4)^2 - 4(1)(5) = -4$ . The quadratic formula asks for a number whose square is $-4$ , and the reals have none; but the complex numbers do, because $(2i)^2 = 4i^2 = -4$ . Feeding $2i$ into the formula in place of the missing square root suggests the candidates

z = \frac{4 \pm 2i}{2} = 2 \pm i.

The formula was derived for real arithmetic, so treat this as a suggestion rather than a conclusion, and let direct substitution give the honest verdict.

Worked example 5 Checking a complex root

Verify that $z = 2 + i$ satisfies $z^2 - 4z + 5 = 0$ .

Compute the square first, expanding and spending $i^2 = -1$ :

z^2 = (2 + i)^2 = 4 + 4i + i^2 = 3 + 4i.

Then assemble the whole left side, distributing $-4$ across $2 + i$ :

\begin{aligned} z^2 - 4z + 5 &= (3 + 4i) - (8 + 4i) + 5 \\ &= (3 - 8 + 5) + (4 - 4)\,i = 0. \end{aligned}

Both the real part and the imaginary part collapse to $0$ , so $2 + i$ really is a solution, certified by nothing but addition and multiplication.

What about the other candidate, $2 - i$ ? It is the conjugate of the root just checked, and the property proved above says you do not need to repeat the computation to trust it. Conjugate both sides of the true statement $z^2 - 4z + 5 = 0$ : the right side, $0$ , is its own conjugate, and conjugation respects both operations on the left side while leaving the real numbers $-4$ and $5$ fixed. So the conjugate of $z$ solves the same equation, meaning $2 - i$ is a root too, with no further arithmetic needed. Roots of a quadratic with real coefficients that miss the real line always arrive in a conjugate pair like this one; the final lesson of this chapter turns that observation into a full proof.

Division undoes multiplication

Worked example 6 Division undoes multiplication

Find the complex number $z$ satisfying $(2 - i)\,z = 4 + 3i$ .

Division exists precisely to undo multiplication, so divide both sides by $2 - i$ , which is legal because $2 - i \ne 0$ :

\begin{aligned} z &= \frac{4 + 3i}{2 - i} = \frac{(4 + 3i)(2 + i)}{(2 - i)(2 + i)} = \frac{8 + 4i + 6i + 3i^2}{2^2 + 1^2} \\ &= \frac{5 + 10i}{5} = 1 + 2i. \end{aligned}

Check it, because a check costs one multiplication:

(2 - i)(1 + 2i) = 2 + 4i - i - 2i^2 = 2 + 3i + 2 = 4 + 3i. \checkmark

Linear equations over the complex numbers solve exactly the way they did over the reals, because the algebraic laws are the same and the reciprocal always exists.

A bit of history (optional)

By the middle of the 1500s, square roots of negatives kept turning up inside perfectly normal problems. Nobody could ban them, since the algebra produced them without being asked. Nobody could use them with any confidence either. One writer had multiplied a pair of them in a single worked case. He set the answer aside as useless. What one such root times another meant in general, nobody had settled.

Rafael Bombelli was an engineer who drained marshes in central Italy for a living. He wrote his algebra book while one of those projects was held up, and it appeared in 1572. What he offered was not so much a discovery as a decision. He treated the impossible root as an object like any other, and built it the first systematic table of signs.

He had no symbol for it, so he used words. A positive multiple of the root he called “plus of minus”, and a negative one “minus of minus”. Then came the rule that opened everything after it: plus of minus, times plus of minus, makes minus. Take the words away and that line is $i \cdot i = -1$ , two centuries before the letter arrived.

The rest followed from ordinary algebra, and Bombelli used it on cubics that had defeated everyone. His key point was a humble one. Once you can multiply, you can expand, and this lesson is that point carried through. You spread the product out as you would any binomial, then trade every $i^2$ for $-1$ .